Tgx·ctgx = 1 ⇒ ctgx = 1/tgx
1/tg²x - 2/tgx = 3
ctg²x - 2tgx = 3
Пусть t = ctgx.
t² - 2t = 3
t² - 2t - 3 = 0
t² - 2t + 1 - 4 = 0
(t - 1)² - 2² = 0
(t - 1 - 2)(t - 1 + 2) = 0
(t - 3)(t + 1) = 0
t = -1; 3.
Обратная замена:
ctgx = -1
x = -π/4 + πn, n ∈ Z
ctgx = 3
tgx = 1/3
x = arctg(1/3) + πk, k ∈ Z
Ответ: x = -π/4 + πn, n ∈ Z; arctg(1/3) + πk, k ∈ Z.